An accurate map of California is spread out flat on a table in Evans Hall, in Berkeley. Prove that there is exactly one point on the map lying directly over the point it represents.
Solution
Identify horizontal position with the Euclidean plane, and let \(C\subset\RR^2\) be the closed region occupied by California.
1. Let\[ T:C\longrightarrow\RR^2 \]send a point \(x\in C\) to the physical horizontal position, on thetable, of the point of the map representing \(x\). Then there is aconstant \(\lambda\) with\[ 0<\lambda<1 \]such that\[ \norm{T(x)-T(y)} = \lambda\norm{x-y} \]for all \(x,y\in C\).
An accurate map differs from the region it represents by a uniform change of scale followed by a rigid motion in the horizontal plane. Thus \[ T(x)=b+\lambda Rx \] for some vector \(b\) and some orthogonal linear map \(R\). The map is strictly smaller than the state, so its scale factor satisfies \(0<\lambda<1\). Orthogonality of \(R\) gives the displayed distance identity.
2. One has\[ T(C)\subseteq C. \]
The physical image \(T(C)\) is precisely the placed map of California. It lies on a table in Evans Hall, hence physically inside California.
3. The map \(T:C\to C\) has a fixed point.
Choose any \(x_0\in C\) and define \[ x_{n+1}=T(x_n). \] Step 2 shows inductively that every \(x_n\) lies in \(C\). By step 1, \[ \norm{x_{n+1}-x_n} = \lambda^n\norm{x_1-x_0}. \] Hence for \(m>n\), \[ \begin{aligned} \norm{x_m-x_n} &\leq \sum_{k=n}^{m-1}\norm{x_{k+1}-x_k}\\ &\leq \frac{\lambda^n}{1-\lambda}\norm{x_1-x_0}. \end{aligned} \] Thus \((x_n)\) is Cauchy. Since \(C\) is closed in \(\RR^2\), it is complete, so \[ x_n\longrightarrow p \] for some \(p\in C\). The similarity \(T\) is continuous, and therefore \[ T(p) = T\left(\lim_{n\to\infty}x_n\right) = \lim_{n\to\infty}x_{n+1} = p. \]
4. The fixed point of \(T\) is unique.
If \(p,q\in C\) are fixed points, then step 1 gives \[ \norm{p-q} = \norm{T(p)-T(q)} = \lambda\norm{p-q}. \] Since \(0<\lambda<1\), this forces \[ \norm{p-q}=0, \] so \(p=q\).
5. A point \(p\in C\) is fixed by \(T\) if and only if the point ofthe map representing \(p\) lies directly over \(p\) itself.
By definition, \(T(p)\) is the physical horizontal location of the map point representing \(p\). Thus \(T(p)=p\) says exactly that these two horizontal positions coincide.
6. Exactly one point on the map lies directly over the point itrepresents.
Existence follows from step 3, uniqueness from step 4, and step 5 identifies fixed points with the points described in the problem.
7. Q.E.D.
Step 6 is the required conclusion.